MATHEMATICS
TARGET: JEE (ADVANCED) 2014 MEDIUM : ENGLISH
CRITICAL QUESTIONS BANK
SUBJECT : MATHEMATICS
MEDIUM : ENGLISH
Contents 1.
Questions
1 - 20
2.
Answer Key
21
3.
Hints & Solutions
22 - 45
QUESTION FORMAT & MARKING CRITERIA SUBJECT : MATHEMATICS A. Questions Format In the booklet check that it contains all the 180 questions and corresponding answer choices are legible. Read carefully the Instructions printed at the beginning of each section. 1. Section 1 contains 53 multiple choice questions. Each question has Four choices (A), (B), (C) and (D) out of which only ONE is correct. 2. Section 2 contains 37 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. 3. Section 3 contains 2 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which Only ONE is correct. 4. Section 4 contains 10 paragraphs each describing theory, experiment, data etc. Twenty questions related to Ten paragraphs with one or two or three questions on each paragraph. Each question of a paragraph has ONLY ONE correct answer among the four choices (A), (B), (C) and (D). 5. Section 5 contains 4 question. Each question contains statements given in two columns which have to be matched. Statements in Column I are labelled as A,B,C and D whereas statements in Column II are labelled as p,q,r,s and t. The answers to these questions have to be appropriately bubbled as illustrated in the following example. 6. Section 6 contains 1 multiple choice questions. Each questions has matching lists. The codes for the lists have coices (A), (B), (C) and (D) out of which ONLY ONE is correct. 7. Section 7 contains 43 questions. The answer to each question is a single-digit integer, ranging from 0 to 9 (both inclusive). 8. Section 8 contains 17 questions. The answer to each question is a double-digit integer, ranging from 00 to 99 (both inclusive). B. Marking Scheme 9. For each question in Section 1, you will be awarded 3 marks if you darken the bubble corresponding to only the correct answer and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark will be awarded. 10. For each question in Section 2, you will be awarded 4 marks if you darken the bubble(s) corresponding to only the correct answer and zero mark if no bubbles are darkened. No negative marks will be awarded for incorrect answers in this section. 11. For each question in Section 3, you will be awarded 4 marks if you darken all the bubble(s) corresponding to only the correct answer(s) and zero mark if no bubbles are darkened. 12. For each question in Section 4 contains 10 paragraphs each describing theory, experiment, data etc. Twenty questions related to Ten paragraphs with one or two or three questions on each paragraph. Each question of a paragraph has ONLY ONE correct answer among the four choices (A), (B), (C) and (D). 13. For each question in Section–5, you will be awarded 2 marks for each row in which you have darkened the bubble corresponding to the correct answer. Thus, each question in this section carries a maximum of 8 marks. There is no negative marking for incorrect answer(s) in this section. 14. For each question in Section 6, you will be awarded 3 marks if you darken all the bubble(s) corresponding to only the correct answer(s) and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark will be awarded. 15. For each question in Section 7, you will be awarded 4 marks if you darken the bubble corresponding to only the correct answer and zero mark if no bubbles are darkened. No negative marks will be awarded for incorrect answers in this section. 16. For each question in Section 8, you will be awarded 4 marks if you darken all the bubble(s) corresponding to only the correct answer(s) and zero mark if no bubbles are darkened. No negative marks will be awarded for incorrect answers in this section.
MATHEMATICS SECTION – 1 : (Only One option correct Type) This section contains 53 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 1.
Let f(x) = sin2(x + ) + sin2(x + ) – 2cos( – ) sin(x + ) sin(x + ). Which of the following is TRUE? (A) f(x) is strictly increasing in x (, ) (B) f(x) is strictly decreasing in x (, ) , and strictly decreasing in x (C) f(x) is strictly increasing in x , 2 2 (D) f(x) is a constant function
2.
3.
4.
If the roots of the equation a(b – c)x2 + b(c – a)x + c(a – b) = 0 (where a, b, c are unequal real numbers) are real and equal and be the roots of equation ax2 + bx + c = 0, a 0 then harmonic mean of is (A) 1 – (B) 1 + (C) – 1 (D) –1 – 1 2 If A = and B = 3 4 is (A) 0
lim+
x 0
e(x
x
-1)
- xx
x
2
(x²) -1
(C) –2
(D) –1
1 8
(C) –1
(D) does not exist
(B)
If the equation 4y3 – 8a2yx2 – 3ay2x + 8x3 = 0 represent three straight lines, two of them are perpendicular then sum of all possible values of a is (A) –
6.
(B) 2
=
(A) 1 5.
a b 3a 3d are two matrices such that AB = BA and c 0, then value of 3b c c d
3 4
(B) 5
(C)
4 5
(D) – 3
If p th , qth , r th terms of an AP are P, Q, R respectively , then which of the following must be CORRECT ? (A) p Q + q R + r P = p R + r Q + q P (B) (p + q + r)th term will be P + Q + R. (C) If p < q < r, then P < Q < R (D) If P,Q,R N then common difference of AP will be an integer..
7.
The set of values of 'a' for which x2 + ax + sin–1(x2 – 4x + 5) + cos–1 (x2 – 4x + 5) = 0 has at least one solution is (A) (– , –
2 ] [ 2 , )
(C) R
8.
(B) (– , –
2 ) ( 2 , )
8 (D) – 4
Let = log10 15 and = log10 16 , a set A = {log 101, log10 2,.........log10 50} the number of elements in set A which can be written in the form of a b c where a,b,c are rational numbers, is (A) 23 (B) 24 (C) 25 (D) none of these
9.
If both the roots of the equation x2 + (3 – 2k) x – 6k = 0 belongs to the interval (–6, 10), then largest value of k is (A) –1 (B) 3 (C) 5 (D) none of these
RESONANCE
Page - 1
MATHEMATICS 10.
1 (A) , 2
11.
(2x 1)
The domain of the function f(x) =
3
(2x 3 x 2 x )
1 (B) , 2 2
sin 1(log 2 x ) is
+
(C) [1, 2]
(D) (1, )
Let G, S, I be respectively centroid, circumcentre, incentre of triangle ABC. If R, r are circumradius and inradius respectively then which of the following is INCORRECT ? (A) SI2 = R2 (1 – cosA cosB cosC) ; A,B,C being angles of triangle. (B) SI2 = R2 – 2Rr (C) SG2 = R2 –
1 2 (a + b2 + c2) ; a, b,c being sides of triangle. 9
(D) SG SI 12.
13.
Two circles are given as x² + y² + 14x –6y + 40 = 0 and x² + y² – 2x + 6y + 7 = 0 with their centres as C1 and C2. If equation of another circle whose centre C3 lies on the line 3x + 4y –16 = 0 and touches the circle C1 externally and also C1C2 + C2C3 + C3C1 is minimum, is x² + y² + ax + by + c = 0 then the value of ( a + b + c) is (A) 2 (B) 0 (C) 16 (D) None of these
[f(x)], If g(x) = 3, f(x) =
x 0, , 2 2 where [x] denotes the greatest integer function and x /2
2(sin x sinn x) | sin x sinn x | ,n R – {1}, then 2(sin x sinn x) | sin x sinn x |
(A) g(x) is continuous and differentiable at x = /2, when 0 < n < 1 (B) g(x) is continuous and differentiable at x = /2, when n > 1 (C) g(x) is continuous but not differentiable at x = /2, when 0 < n < 1 (D) g(x) is neither continuous nor differentiable, at x = /2, when n > 1
14.
If
sec 2 x 2010
sin
2010
x
dx =
P( x ) (sin x )
2010
+ C, then value of P is 3
1 (A) 0
15.
(B)
(C)
3
(D)
3
3 3 2
is equal to If ˆi ˆj bisects the angle between c and j k , then value of c.j (A) 0
1
(B)
(C)
2
1 2
(D) 1
16.
If A and B are two square matrices such that B = – A–1 BA, then (A + B)2 is equal to (A) O (B) A2 + B2 (C) A2 + 2AB + B2 (D) A + B
17.
The value of where = sin–1
(A) 0
18.
(B)
4
2 3 + cos –1 4
12 + sec –1 ( 2 ) is equal to 4 (C)
6
(D)
2
x , xQ f(x) = . The number of points of continuity of f(x) is cos x , x Q
(A) 0
RESONANCE
(B) 1
(C) 2
(D) infinitely many
Page - 2
MATHEMATICS 2
19.
Let f(x) =
x
2
dy 1 y3
. The value of the integral
0
(A) 1
20.
xf(x) dx is equal to
(B)
4 3
(C)
2 3
Let P be a point on ellipse 4x² + y² = 8 with eccentric angle
(D)
1 3
. If tangent at P intersects the x-axis at A and 4
y-axis at B and normal at P intersect the x-axis at A' and y-axis at B'. The ratio of area of triangle APA' to area of triangleBPB' is (A) 1 : 1
21.
(B) 2 : 1
1 sin x
1 cos x (A) e–x/2 sin
22.
(D) 4 : 1
5 9 e–x/2 dx, x , 2 2
x +C 2
(B) – e–x/2 sec
x +C 2
(C) –e–x/2 sin
x +C 2
(D) –e–x/2 cos
x +C 2
e [ x ] |x| – 2 = (where [.] denotes greatest integer function) [ x ] | x |
lim x 0 –
(A) 2 –
23.
(C) 3 : 1
1 e
(B) –1
If tan1(1)
(C) 2 – e
(D)
1 –2 e
1 1 1 1 1 1 1 tan1(2) tan 1(3) , tan (1) 2tan 3tan and 2 3 2 3
p r cot 1(3) , q s
where p, q, r, s N and are in their lowest form then which of the following is INCORRECT ? (A) p r 0 (B) q 4s (C) p q r s 42 (D) pr 1 q 24.
The length of sub-normal at any point P(x,y) on the curve, which (curve) is passing through Q(0, 1) is unity. The area bounded by the curves satisfying this condition is equal to (A)
2 3
(B)
4 3
(C)
1 3
(D)
8 3
25.
Let f(x) = x + x² + x4 + x8 + x16 + x32 + ......the coefficient of x10 in f(f(x)) is (A) 28 (B) 40 (C) 52 (D) none of these
26.
If 1, 1, 2, 3,.....2008 are (2009)th roots of unity, then the value of
2008
(A) 2009
27.
Let lim x
2
(B) 2008
(cos x ) . sin(sec x ) 2
(C) 0
r(
r
2009–r ) is equal to
r 1
(D) – 2009
, where , R
x – 2
(A) = 2, = 0
RESONANCE
(B) = 0, = 2
(C) =
7 ,=0 3
(D) =
5 ,=0 3
Page - 3
28.
MATHEMATICS Let A(x1, y1), B(x2, y2) C(x3, y3) be vertices of triangle ABC with BC = a, AB = c, AC = b. If algebraic sum of 3ay 1 3by 2 3cy 3 3cx 3 3ax 1 3bx 2 , , , to a , M , N perpendicular distances from L a b c a b c a b c a b c a b c a bc variable line is zero then all such lines passes through (A) orthocentre of ABC (B) centroid of ABC (C) circumcentre of ABC (D) incentre of ABC
29.
If |f(x)| 1 x R and f(0) = 0 = f(0) then which of the following can be TRUE? 1 1 (A) f 3 5
30.
1 3
(D) none of these
1 4
(B)
1 2
(C) 1
(D)
7 8
The principle argument of z = x + iy, if it lies in second quadrant is equal to (A)
32.
(C) f – 3
If f(x) = cos 8{x} + sin 2x cosec 2x (where {.} represents fractional part function), then fundamental period of f(x) is (A)
31.
1 1 (B) f – 3 4
y tan –1 2 x
–1 (B) – tan
y x
–1 (C) tan
y x
(D)
y – tan –1 2 x
x 2 {e1/ x } , x 0 If the function f(x) = , x 0 where {.} denotes fractional part function, is continuous at x = 0, then k (A) k = 1 (B) f(x) is non-derivable at x = 0 (C) f(x) is derivable at x = 0 (D) f(x) is continuous at every point in its domain. 3
33.
f(x) = – x2 + 1, g(x) = – (A) odd function
34.
If a = 2 , then sum of series cot–1(2a–1 + a) + cot–1 (2a–1 + 3a) + cot–1(2a–1 + 6a) + cot–1 (2a–1 + 10a) + .... upto infinite terms, is (A)
4
x . Then (gofogofogogog)(x) is (B) even function (C) polynomial function (D) identity function
(B)
2
(C)
3
(D)
6
35.
The normal to the curve 5x5 – 10x3 + x + 2y + 6 = 0 at P (0, –3) meets the curve again at two points, then the equations of the tangents to the curve at these points is/are (A) 2x – y – 3 = 0 (B) 2x + y – 3 = 0 (C) 2x + y + 3 = 0 (D) 2x – y + 3 = 0
36.
If f(x + y + 1) = (A) 1 – x
37.
f(x) =
f (x)
2
f (y)
2
and f(0) = 1, x, y R then f(x) can be
(B) 1 – x
(B) f is injective (D) f –1(x) = f(x) have no solution.
A normal chord drawn to a parabola y2 = 4x at any point P intersects the parabola again at a point Q, then the minimum distance of Q from the vertex of the parabola is equal to (A) 4 2
39.
(D) x2 – 1
cos –1(e 2 x sin –1 x )
(A) domain of f is [–1, 1] (C) range of f is [0, 5] 38.
(C) (x + 1)2
(B) 4 3
(C) 4 5
(D) 4 6
Let p(x) be a real polynomial of least degree which has a local maximum at x = 2 and a local minimum at x = 4. If p(2) = 8, p(4) = 1 , then p(0) is (A)
168 5
RESONANCE
(B) 42
(C) 43
(D) 45 Page - 4
MATHEMATICS 40.
41.
2 . If g(x) is the function whose graph is the reflection of the graph of f(x) 3 with respect to line y = x, then g(x) equals to Let f(x) = (3x + 2)2 – 1, – < x –
(A)
1 (–2 x 1), x –1 3
(B)
1 (–2 – x 1), x –1 3
(C)
1 (–1– x 2 ), x –2 3
(D)
1 (–1 x 2 ), x –2 3
9 33 129 + cot–1 ............ is cot–1 + cot–1 2 4 8
(A)
42.
4
100 + The sum 98
(B)
44.
101 (B) 99
47.
48.
n = n Cr r
100 (C) 99
101 (D) 98
(C) (1, 2]
The solution set of the inequality (cosec–1x)2 – 2(cosec–1x) (B) 2
(D) (–, 2]
cosec–1x – is (–, a] [b, ), then 6 3
(C) – 3
(D) 1
1+ Z1 + Z 2 + Z3 ......... + Z7 = 1+ Z8 + Z9 + Z10 + ....... + Z14
arg
(A)
1 arg (Z1 Z2 .........Z7) 2
(B) arg (Z1 Z2 .........Z7)
(C)
1 arg (Z1 Z2 .........Z7) 4
(D)
1 arg (Z1 Z2 .........Z7) 8
2 2 Image of the curve xy = 1 in the curve x x 1 y y 1 = 1 (on coordinate plane) is (A) xy = 1 (B) xy + 1 = 0 (C) xy = 0 (D) x2 + y2 = 1
|1 + Z1 + Z2 +.....+ Z7 | = 7 (A) sec 15
49.
(D) none of these
4 x – x 3 n(b 2 – 3b 3) , 2 x 3 Let f(x) = . Find all the possible real values of b such that f(x) has the x – 18 , x3
(a + b) is equal to (A) 0
46.
4
A circle with centre (3, 3) and of variable radius cuts the hyperbola x2 – y2 = 36 at the points P,Q,R and S. If the locus of centroid of PQR is (x – 2)2 – (y – 2)2 = , then the value of is (A) 3 (B) – 2 (C) 4 (D) – 3
smallest value at x = 3. (A) (–, 2] [3, ) (B) (–, 1] [2, )
45.
(C)
99 98 3 2 + +.......+ + is equal to, where 97 96 1 0
100 (A) 98
43.
2
(B)
1 7 sec 2 15
7 (C) 2 sec 15
(D) None of these
Let O be centre, S, S be foci of hyperbola. If tangent at any point P on hyperbola cuts asymptotes at M and N then OM + ON = (A) |SP – SP| (B) SP + SP (C) SS (D) distance between vertices
RESONANCE
Page - 5
50.
MATHEMATICS DABC be a tetrahedron such that AD is perpendicular to the base ABC and ABC = 30º. The volume of tetrahedron is 18. If value of AB + BC + AD is minimum then length of AC is (A)
51.
3
6 3
6 2
(C)
2
6 3
(D)
2
6 2
(B) (–, 0]
(C) [0, )
(D) (–, –1) (1, )
If g(x) = 2f (2x3 – 3x2) + f (6x2 – 4x3 – 3), x R and f (x) > 0, x R, then g(x) is increasing on the interval 1 (A) , 0,1 2
53.
3
1– x2 , is The solution set of values of x satisfying equation 2cot–1x + cos–1 1 x2 (A) all real numbers
52.
(B)
1 (B) ,0 1, 2
(C) 0,
(D) ,1
Let a circle with centre at C be made to pass through the point P(1,2), touching the straight lines 7x – y = 5 and x + y + 13 = 0 at A and B respectively. If tangents at A and B intersect at Q, then (A) area of quadrilateral ACBQ can be 100 sq. units
(B) radius of circle can be
(C) area of quadrilateral ACBQ can be 200 sq. units
(D) radius of circle can be10
40
SECTION – 2 : (One or more options correct Type) This section contains 37 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.
54.
Consider the locus of the complex number z in the Argand plane is given by Re(z) –2 = |z – 7 + 2i|. Let P(z1) and Q (z2) be two complex number satisfying the given locus and also satisfying
z - (2 + i) arg 1 = ( R) then the minimum value of PQ is divisible by z 2 - (2 + i) 2 (A) 3 (B) 5 (C) 7 (D) 2 55.
Let f(x2 + y) = (f(x))2 + f(y) for all x, y R, then (A) f(x) is odd (B) f(x) is even f(x) = 0 x R (C) f(x) is continuous at x = 0 it is continuous everywhere (D) f(x) is differentiable at x = 0 f(x) = x f (0), x R
56.
Let f(x) = x3 + px2 + qx + r, where p,q and r are integers, f(0) and f(–1) are odd integers. Which of the following is/are CORRECT ? (A) f(1) is an even integer (B) f(1) is an odd integer (C) f(x) = 0 has three distinct integer roots (D) f(x) = 0 cannot have three integer roots.
57.
Let a,b,c R. If ax 2 + bx + c = 0 have one root < –1 and other root > 1, then (A) 1 –
58.
b c 0 a a
b c 0 a a
The value(s) of x satisfying 1 – log9(x + 1)2 = (A) 1
59.
(B) 1 +
(B) –2
(C) 1 –
b c 0 a a
1 log x 5 is/are 3 x3 2 (C) –7
(D) 1 –
b c 0 a a
(D) –4
x m f ( x ) h( x ) 3 If g(x) = mlim when x 1 and g(1) = e3 such that f(x), g(x) and h(x) are continuous 2x m 4 x 1 functions at x = 1, then (A) f(1) = 2e3 (B) h(1) = 5e3 – 3 (C) f(1) + h(1) = 7e3 + 5 (D) f(1) – h (1) = 7e3 + 5
RESONANCE
Page - 6
60.
MATHEMATICS A parabola C whose focus is S(0,0) and passing through P(3, 4). Equation of tangent at P to parabola is 3x + 4y – 25 = 0. A chord through S parallel to tangent at P intersects the parabola at A and B. Which of the following are CORRECT ? (A) Length of AB is 20 units (B) Latus rectum of parabola is 20 units (C) Only one real normal can be drawn from the point (–3, –4) (D) Only one real normal can be drawn from the point (–6, –8) 11 , where a and b are natural numbers then b is divisible by 4 + 8 - 32 + 768 = a 2 cos a b (A) 2 (B) 4 (C) 3 (D) 6
61.
If
62.
For the function f(x) = (x2 + bx + c) ex and g(x) = (x2 + bx + c) ex + ex (2x + b), which of the following holds good ? (A) f(x) > 0 for all real x g(x) > 0 (B) f(x) > 0 for all real x g(x) > 0 (C) g(x) > 0 for all real x f(x) > 0 (D) g(x) > 0 for all real x f(x) > 0
63.
1 cos If the equation 2 1 x
of 'a' are (A) (–3, 1)
64.
2
1 1 cos1 x a 2 a2 0 has only one real solution then subsets of values 2
(B) (–, –3]
(C) [1, )
(D) [–3, )
x2 , x 2 2 , 2x3 f(x) = x 3 , x3 x – 18 (A) f(x) is continuous in R. (B) f(x) is discontinuous at 'a' for some a R. (C) f(x) is continuous at infinitely many real x. (D) f(x) is discontinuous at infinitely many real x. 15
65.
4 1 Which of the following must hold good for the expansion of the binomial x 3 x (A) There exist a term which is independent of x. (B) 8th and 9th terms of the expansion have the greatest binomial coefficient (C) Coefficients of x32 and x–17 are equal (D) If x =
66.
2 , then number of rational terms in the expansion is 5
Which of the following is NOT in the solution set of the inequality |x 2 – 1| + |2x 2 – 3| < |2 – x 2| ? (A) (–1, 1)
67.
(B) (–2, 2)
1 (C) – 1 , 2
(D) (3, )
1 1 Let a function f(x), x 0 be such that f (x) f f (x) f then f(x) can be : x x (A) 1 – x2013 (C)
68.
?
2 tan 1 | x |
(B) (D)
| x | 1 2 , k being arbitrary constant 1 k n | x |
TP and TQ are tangents to parabola y2 = 8x and normals at P and Q intersect at a point R on the parabola. The locus of circumcentre of TPQ is a parabola whose (A) Vertex is (2, 0)
(C) length of latus rectum is 1
RESONANCE
7 (B)foot of perpendicular from focus on the directrix is , 0 4 9 (D) focus is , 0 4
Page - 7
69.
MATHEMATICS The function f(x) = 2|x| + |x + 1| – ||x – 3| – 3|x|| has a local minimum or a local maximum at x = (A) 0
70.
(B) –
3 2
(C)
3 4
(D) 3
If z1 , z2 ,z 3 , z4 are roots of the equation a0z4 + a1z3 + a2z2 + a3z + a4 = 0, where a0 , a1 , a2 , a3 and a4 are real, then (A) z 1 , z 2 , z 3 , z 4 are also roots of the equation (B) z1 is equal to at least one of z 1 , z 2 , z 3 , z 4 (C) – z 1 , – z 2 ,– z 3 ,– z 4 are also roots of the equation (D) None of the above.
71.
sin(e x–3 – 1) = c, which of the following is/are CORRECT ? X 3 n(x – 2) lim
2 (C) c is a positive integer (A) sin–1c = –
72.
(B) c = 0 (D) c is neither prime nor composite
Let a, b and c be three unit vectors such that | a b c | = 3 and (a b).(b c ) (b c ).(c a) (c a).(a b ) . Which of the following are CORRECT ? (A) The maximum value of is 0
(B) If is maximum then the volume of parallelepiped determined by a, b and c is
3
(C) If is maximum then the value of (2a 3b 4c ).(a b 5b c 6c a) is 32. (D) None of these 73.
Let g be the inverse of the continuous function f, Let there be a point (, ), where , is such that it satisfies each of y = f(x) and y = g(x) then (A) the equation f(x) = g(x) has infinitely many solutions (B) the equation f(x) = g(x) has atleast 3 solutions (C) f must be a decreasing function of x (D) g can be an increasing function of x
74.
The maximum and minimum value of ab sin x + b 1 a 2 cos x + c (|a| 0), lie in interval (A) [c – b, b + c] (B) (b – c, b + c) (C) [c – 2b, c + 2b] (D) [–c, c]
75.
2 If lim (log 3 (ax 3x 1))
76.
x2
log ( x 1) 3
, where is a finite real number then
(A) a = – 1
(B) 'a' can have more than one values
(C) = e–2/3
(D) = e–1/3
Which of the following functions is an injective (one-one) function in their respective domain? (A) f(x) = 2x + sin 3x (B) f(x) = x. [x],(where[.] denotes the G.I.F.) (C) f(x) =
2x +1 4 x -1
RESONANCE
(D) None of these
Page - 8
77.
78.
MATHEMATICS The centre of a circle S = 0 lies on 2x – 2y + 9 = 0 and S = 0 cuts orthogonally the circle x2 + y2 = 4. Then the circle must pass through the point (A) (1, 1) (B) (– 1/2, 1/2) (C) (5, 5) (D) (– 4, 4)a
1 1 x If 2f(x) + xf - 2f 2sin x + = 4cos² + xcos , x R - {0} then which of the following x 4 2 x statements(s) is/are true? 1 (A) f(2) + f = 1 2
(B) f(2) + f(1) = 0
1 (C) f(2) + f(1) = f 2
1 (D) f(1) + f = 1 2
11 , where a and b are natural numbers then (b – a) is divisible by 4 + 8 - 32 + 768 = a 2 cos b (A) 2 (B) 23 (C) 69 (D) 46
79.
If
80.
Equation of the plane passing through the line of intersection of the two planes r . n1 q1 and r . n2 q2 and parallel to the line of intersection of r . n3 q3 and r . n4 q4 is (A) dependent on n1 . n 3 (B) dependent on n3 . n4 (C) independent of q1 and q2 (D) independent of q3 and q4
81.
Let ABCD be a tetrahedron, where A = (2,0,0), B = (0,4,0). If edge CD lies on line
that centriod (,,) of tetrahedron satisfies
x 1 y 2 z 3 , such 1 2 3
5 2 – y1 – z1 then which of the following are 1 a b
2 –
CORRECT? (A) a + b =
5 2
(B) y1 + z1 =
19 4
(C) y1 – z1 =
1 4
(D) a + b + y1 = 5
82.
The line 3x + 6y = k intersects the curve 2x2 + 2xy + 3y2 = 1 at points A and B. The circle on AB as diameter passes through the origin. The possible value of k is (A) 3 (B) 4 (C) – 4 (D) –3
83.
A rod of length 2 units whose one end is (1, 0, –1) and other end touches the plane x – 2y + 2z + 4 = 0, is rotated on this plane, then (A) the rod sweeps a solid structure whose volume is cubic units (B) the area of the region which the rod traces on the plane is 2 (C) the length of projection of the rod on the plane is
3 units 2 2 5 , 3 3 3
(D) the centre of the region which the rod traces on the plane is ,
84.
Let f : 0, [0, 1] be a differentiable function such that f(0) = 0, f = 1, then 2 2
(A) f () =
1 ( f ( ))2 for all 0, 2
(C) f() f () =
1 for at least one 0, 2
RESONANCE
(B) f () =
(D) f () =
2 for all 0, 2
8 2
for at least one 0, 2
Page - 9
85.
86.
MATHEMATICS Two equal sides of an isosceles triangle are given by the equation 7x – y + 3 = 0 and x + y – 3 = 0 and its third side passes through the point (1, – 10). The equation of the third side can be (A) x + 3y + 29 = 0 (B) x – 3y = 31 (C) 3x + y = 3 (D) 3x + y + 7 = 0 Let f(x) = cos–1 (cos2x) and g(x) = |cos x| then (A) number of solution of f(x) = g(x) in [0,2] is 4. (B) max {f(x), g(x)} is a periodic function (C) max {f(x), g(x)} is a non differentiable function for some x, (D) min {f(x), g(x)} is an even function
87.
3 Consider f(x) = 2sinx + sin 2x , x 0, 2
3 3 2 (D) Least value of f(x) is – 2
(A) Greatest value of f(x) is 3
(B) Greatest value of f(x) is
(C) Least value of f(x) is zero 88.
For a Function f : A B such that n(A) a, n(B) b (a,bN) then which of the following statements s must be CORRECT ? (A) If function is one - one, onto, then a>b (B) If function is one - one, into, then ab (D) If function is many - one, into, then a 0 for
sin 1(log2 x ) , 0 log2 x 1
x [1, 2] ........(2)
RESONANCE
Page - 22
MATHEMATICS So by (1) and (2) we get x [1, 2]
1
(sin x)
1 3 11. R2 – SG2 = (a2 + b2 + c2) (abc)2/3 .........(1) 9 9 abc = 4R = 4Rrs = 4Rr
dx = I1 – I2
2010
Applying by parts on
abc 2Rr(abc)1/3.3 2
tan x
I1, we get
2010
(sin x )2010
tan x cos x
(sin x )
2011
dx
(abc)2/3 6Rr
1 R – SG 6Rr 3 2
=
2
SG2 R2 – 2Rr SG2 SI2 SG SI
tan x (sin x )
2010
2010
tan x
12. Image of the centre C2 (1, –3) in the line 3x + 4y –16 = 0 is P (7, 5). Now for C1C2 + C2C3 + C3C1 to be minimum C1, C3 and P should be on same line so C3 = (0, 4) C3 = (0, 4)
15.
I1 – I2 =
P(x) = tan
2
x (x y) y
(sin x )2010
+ c P(x) = tanx
2
1 2
2(4x 2 y 2 4xy) 2x 2 2y 2 2xy
50 5 2
6x 2 6xy 0
3 2
so radius of C3 = 2 2 Equation of C3 (x – 0)² + (y – 4)² = 8 x² + y²– 8y + 8 = 0, a = 0, b = –8, c = 8
x y c x (i k)
x = 0 or
13. For 0 < n < 1, sin x 1, sin x > sinn x Now, for 0 < n 2 28. Let variable line be x + my + n = 0
tan1 1 tan1 2 tan1 3 1 tan 2
2
(cos x)
3ax1 3amy1 n a b c a bc P1 = 2 m2 3bx 2 3bmy 2 n a b c a b c P2 = 2 m2
1 1 tan1(1) tan1 tan1 2 3 2
RESONANCE
Page - 24
MATHEMATICS 3cx 3 3cmy 3 n abc abc P3 = 2 m2
(n 1)a na 2 2 = tan-1 (n 1)a na 1 2 . 2
Given P1 + P2 + P3 = 0
3(ax1 bx 2 cx 3 ) 3m(ay1 by 2 cy 3 ) + (a b c ) (a b c )
+ 3n = 0
= tan-1 (n 1)
(ax1 bx 2 cx 3 ) m(ay1 by 2 cy 3 ) + +n=0 (a b c ) (a b c )
Hence line
x + my + n = 0 passes through incentre
ax 1 bx 2 cx 3 ay 1 by 2 cy 3 , of triangle ABC. abc abc
a na – tan–1 2 2
Put n = 1, 2, 3, .........., n we have
Sn = tan–1
a (n 1)a a a –tan–1 ; S = – tan–1 = cot-1 2 2 2 2 2
a=2
S =
29. According to LMVT
35. 5x5 – 10x3 + x + 2y + 6 = 0 ......... (i) Differentiating w.r.t. x
f ( x ) – f (0 ) = f (C1 ) 1 x
25x4 – 30x2 + 1 + 2
f(x) | x| f ( x ) – f (0) = f (C2 ) | x | x
again by LMVT f(x)
x2
30. f(x) = cos 8 {x} = cos (8 x – 8 [x]) = cos 8x. Its period is
1 . 4 Period of sin 2x cosec2x is
1 1 period of f(x) is 2 2
32. f(0) = k and xlim f(x) = xlim x2{e1/x} = 0 0 0
2 1/ h lim h {e } = 0 h 0 h
f(x) is derivable at x = 0 and f (0–) =
2 –1/ h } lim h {e =0 h 0 –h
due to {e1/x}, will be discontinuous at all value of x, where e1/x becomes an integer. i.e. at x = log2e, log3e,.....etc. 33. f is even and g is odd gogog is odd
dy 1 1 – – dx (0 ,–3 ) = 2 ( 0 – 0 + 1) = 2 equation of the normal at P(0,–3) is P(0,–3) y + 3 = 2 (x–0) y = 2x – 3 .......... (ii) solving equation (i) and (ii) 5x5 – 10x3 + x + 4x – 6 + 6 = 0 5x (x4–2x2+1) = 0 5x (x2–1)2 = 0 x=0,1,–1 y=–3,–1,–5 Normal at point P(0,–3) meets the curve again at two points (1,–1) and (–1,–5) . Now equation of tangent at point (1,–1) is
fogogog is even
fogogog
gofogofogogog is even.
gofogofogogog
n(n 1) 34. nth term of 1 + 3 + 6 + 10 + ....... is n = 2
1
dy
y + 1 = dx (x–1) (1,–1) y + 1 = 2 ( x –1) 2x – y – 3 = 0 ........ (iii) Equation of tangent at point (–1, –5) is
y+5=
4 n(n 1)a 2 n(n 1)a = cot –1 2a 2
y + 5 = 2 (x + 1)
and f(x + y + 1) =
2x – y – 3 = 0
........ (iv)
a 2 2a –1 –1 =tan na (n 1)a 2 = tan 4 n ( n 1 ) a 1 2 2
f (x)
f (y)
2
substituting x = 0, y = 0, we get f(1) = (1 + 1)2 = 4 substituting y = 0, we get
f ( x 1)
2
f (x) 1
f(0) = 1,f(1) = 22, f(2) = 32, f(3) = 42 and so on hence f(x) = (1 + x)2 37. Composition of one - one function is one - one 38.
RESONANCE
dy (x+1) dx (–1,–5 )
36. f(0) = 1 gogog
T n = cot –1 2a
dy =0 dx
dy 1 = – (25x4 – 30x2 + 1) dx 2
for f(x) to be continuous f(0) = xlim f(x) k = 0 0 f (0+) =
4
AQ =
a2 t 24 4a2 t 22 = at2
t 22 4 = t 2 t 22 4
( a = 1) Since t2 = – t1 –
–2 2 = (– t1) + t1 t1 Page - 25
MATHEMATICS t1 < 0 (as shown is diagram
x1 = 3(h – 2) and y1 = 3(k – 2) orthocentre (x1, y1) lies on x2 – y2 = 36 9(h – 2)2 – 9(k – 2)2 = 36 (h – 2)2 – (k – 2)2 = 4 locus of centroid G(h, k) is (x – 2)2 – (y – 2)2 = 4 = 4
now apply A.M. G.M
44. Here f (x) =
–2 (–t1 ) t1 2 2
(–t1) +
lim
= t2
2 2
84 = 4 6
2 2
39. p = (x – 2) (x – 4) = (x2 – 6x + 8) p(x) = (x3/3 – 3x2 + 8x) + 40. y = f(x) = (3x + 2)2 – 1, – < x –
x 3 –
y + 1 = (3x + 2)
3x + 2 = ±
x3
,
f(x) f(3) i.e. – 15 + n(b2 – 3b + 3) – 15
i.e. n (b2 – 3b + 3) 0 b2 – 3b + 3 1 i.e. i.e. b (–, 1] [2, ).
t2 – 2t
2 3
y 1
b2 – 3b + 2 0
45. Let cosec–1x = t
(t – 2) 6
(t – 2)(t – /6) 0 t /6 or t 2 (not possible) t /6 0+ < t /6 or –/2 t < 0– 0+ < cosec–1x /6 or –/2 cosec–1x < 0– 2 x < or – < x –1 x [2, ) (– , –1] a = –1, b = 2 a+b=1
1 Z1 Z2 ......... Z7 1 Z8 Z9 Z10 ....... Z14
It is clear from the graph that if x (–, –2/3], then y [–1,) 2
2x3
f(x) is decreasing in [2, 3) and increasing in [3, ). Now, f(x) will have least value at x = 3 if
–2 2 2 t1
(AQ)min =
4 – x 2 1
46. arg
( x –2/3, negative sign) 3x + 2 = –
y 1 = arg
x=
3
g(x) =
41.
–2 – y 1
cot 1
cot 1
1 Z1 Z 2 ......... Z7 1 Z1 Z2 .......... Z7
= g(y) = f–1(y) = 2 arg (1+Z1 + Z2 + .........+Z7)
–2 – x 1 3
Z arg = 2 arg (Z) Z
9 2 42 tan 1 tan 1 tan 1 4 tan 1 2 2 9 1 4.2
arg (Z3+Z4)
33 4 84 tan 1 tan 1 tan 1 8 tan 1 4 4 33 1 4.8
z5
z4
z3 z2 z1
z6
129 16 8 8 cot 1 tan1 tan1 16 tan1 8 tan1 8 1 16.8 129 Similarly Tn = sum = as n
z7
1
tan 1 2n1 tan 1 2n on adding we get
tan 1 2n 1 tan 1 2
sum =
– tan–1 2 = cot–1 2. 2
42. Let S = 2C0 + 3C1 + 4C2 + 5C3 +.......+ 99C97 + 100C98 S = 3C0 + 3C1 + 4C2 + 5C3 +.......+ 99C97 + 100C98 S = 4C1 + 4C2 + 5C3 + .......+ 99C97 + 100C98 S = 5C2 + 5C3 + .......+ 99C97 + 100C98 S = 101C98
= 2 arg (Z3 + Z4) =2x
7 14 15 15
Also arg (Z1 Z2 Z3.......Z7) =
2 4 6 14 .......... 15 15 15 15
=
2 15
43. Orthocentre lies on the rectangular hyperbola and
h=
(1+ 2 +3 +.......+7)
56 15
2(3 ) x1 2(3) y1 and k = 3 3
RESONANCE
Page - 26
MATHEMATICS 1 x2 1 = y y2 1
47. x +
y=–
1 take loge n ( x x 2 1) n 2 y y 1
x2 1 +
OM + ON = ae(2sec) = 2ae(sec) SP + SP = e(asec –
y2 1
.....(1)
1 2
y 1 =
y2 1 + y =
x+y=–
a 2 b 2 (sec + tan) = ae(sec + tan)
ON =
y2 1 – y
x2 1 + x =
similarly y +
b x .......(3) a
M [a(sec – tan), b(sec – tan) ] and N [a(sec + tan), b(sec + tan) ]
y2 1 y = n 2 2 y 1 y
x+y=
b x ..........(2) a
y=
x2 1 –
a a ) + e(asec + ) = 2aesec e e
D 50.
1 3
V =
x x2 1
AD (
1 2
x2 1 – x y2 1
.....(2)
B 90°
adding (1) & (2) x + y = 0 Image of xy = 1 in xy = 1 it self.
30°
A
48. |1+ Z1 + Z2 + Z3 + ..........Z7| = | 1 + Z7| + |Z1 + Z6| + |Z2 + Z5| + |Z3 + Z4|
7 5 3 + 2cos + 2cos + 2cos 15 15 15 15
= 2cos
=2
4 3 4 2 cos 15 cos 15 2 cos 15 cos 15 3 cos 15 cos 15
4 = 4 cos 15 = 8 cos
=
15
cos
2 15
cos
4 15
=
1 7 sec 2 15
8 sin(8 / 15) 8 sin( / 15)
C AB. BC sin 30°)
AD. AB. BC = 216 Now, AB + BC + AD 3 (AD. AB. BC) AB + BC + AD 18
1/3
and minimum value occurs when AB = BC = AD = 6 Hence AC =
=3
6 2
(x – 1)2
a2
y2 –
b2
1
Tangent at P(asec, btan)
x y sec tan 1 ........(1) a b Asmptotes are
RESONANCE
52. Since f(x) > 0 f (x) is always increasing. g(x) = 2f (2x3 – 3x2) × (6x2 – 6x) + f (6x2 – 4x3 – 3) (12x – 12x2) = 12(x2 – x) . (f (2x3 – 3x2) – f (6x2 – 4x3 – 3)) = 12x (x – 1) [f (2x3 – 3x2) – f (6x2 – 4x3 – 3)] For increasing g(x) > 0 Case-I x < 0 or x > 1 f (2x3 – 3x2) > f (6x2 – 4x3 – 3) 2x3 – 3x2 > 6x2 – 4x3 – 3 { f(x) is increasing}
49.
x2
AB2 BC 2 2AB BC cos 30
1 1 x > 0 x > – 1 x – , 0 (1, ) 2 2 2
Case-II If 0 < x < 1 f (2x3 – 3x2) < f (6x2 – 4x3 – 3) (x – 1)2
1 x < 0 2
x0 x>–1 3(x + 3) = (x + 1)(x + 5) x2 + 3x – 4 = 0 x = – 4 or x = 1 x = – 4 rejected ( x > – 1) x=1 Case - II x+1 0 iff D=b2 – 4c + 4=D + 4 < 0
63.
2 cos1 x
let
2
2
cos1 x
1 a 2 2
=t
cos1 x
Let circumcentre of TPQ be O(h,k) O(h, k) will be mid point of RT
a2 0
h=
at1t 2 at 32 h = 2 + t32 2
t2 and k =
Now equation
1 t2 a t a2 0 2
has one root 2 or
........ (1)
a(t1 t 2 ) 2at 3 k = t3 2
.........
(2) h = 2 + k2 locus of O(h, k) is O(h, k) y2 = x – 2 y2 = 1(x – 2)
greater than 2 and other root less than 2
ƒ(2) 0 1 2 4 a 2 2 a 0
4 2a 1 a2 0 a2 2a 3 0 (a 3)(a 1) 0 a 3
or
a 1
69.
64. f is discontinuous at points outside its domain. 65.
Tr+1 15Cr(x4)(15–r)(x–3)r = 15Crx60–7r (A) for the term independent of x, 60 – 7r = 0 r is not an integer there is no term independent of x. (B) n = 15 is odd
n–1 n 1 nCr will be maximum if r = or r = 2 2 i.e. r = 7 or r = 8 binomial coefficient of 8th and 9th terms will be greatest (C) for the cofficient of x32; 60 – 7r = 32 r = 4 cofficient of x32 is = 15C4 for the coeff. of x–17; 60 – 7r = – 17; r = 11 coeff. of x–17 is = 15C11 = 15C4 (C) is correct (D) If x =
2 , Tr+1 = 15Cr 2
70. a0z4 + a1z3 + a2z2 + a3z + a4 = 0 Taking conjugate on both sides. a0 ( z )
4
+ a1 ( z )
3
+ a2 ( z )
2
+ a3 z + a4 = 0
z1, z 2 , z 3 , z 4 are the roots of the equation if z1 is real,
then
z1 is also real and if z1 is non real, then z1 is also root
because imaginary roots occur in conjugate pair.
sin(e x–3 – 1) X 3 n(x – 2)
71. lim
put x = 3 + h
60–7r 2
RESONANCE
Page - 29
MATHEMATICS h
h
sin(e – 1).(e 1) h sin(e – 1) = lim h h 0 h0 n(1 h) n(1 h) (e – 1)h
lim
=1×1×1 =1
72.
ab . bc bc . ca ca . a b
b.c a.c a.b
2 1 1 is one-one function. 4x 1 2x 1
c=4 Moreover – 2g + 2f + 9 = 0 ( (– g, – f) satisfy the given equation) S x2 + y2 + 2gx + 2fy + 4 = 0 x2 + y2 + (2f + 9)x + 2fy + 4 = 0 (x2 + y2 + 9x + 4) + 2f (x + y) = 0 It is of the form S + P = 0 and hence passes through the intersection of S = 0 and P = 0 which when solved give (– 1/2, 1/2), (– 4, 4).
a b c 3 a b c . a b c 3 a.b b.c c.a 0 a.b . b.c b.c c.a c.a a.b a.b b.c c.a a.b . b.c b.c c.a c.a a.b
1 -2f(1) 2
78. Replace x by 2, 2f(2) + 2f
1 =2 + f(1)-----(1) 2
Solve (1) and (3) => f
73. As the point (, ), lies on both f(x) and g(x) , the point () will also lie on both the curves and as the functions are continuous they must cross (meet on) the line y = x in between. f must be on decreasing path, for all these to happen.
79. L et
= cos–1a. So f(x) = 0 x =(2m +1)
2
Also f(x) = –b sin (x+). Thus f < 0 and f 2
1 = 0 ; 2
3 > 0 2
f(2) = 1
1 6 3 3 2 co s 3 2 6
So
4
8
4 4 sin
3 2 3 2 co s 24
6
1 1 4 1 co s 24 S o a 2, b 4 8
4
4 4 co s
8 8 co s
12
1 1 24
4 .2 co s ²
..............(3)
7 6 8 3 2 co s
co s
1 a 2 sin x = b cos (x + )
..............(2)
1 1 1 5 , 2f + f(2) + 2 = 2 2 2 2
Replace x by
=4
f(2) + f
Replace x by 1, f(1) = -1
0[since , x y z 0, xy yz zx 0] max 0only when a.b b.c a.c 0 a b, b cand c a 2a 3b 4c . a b 5b c 6c a 10a.(b c) 18b.(c a) 4c.(a b) 32
Where
f (x)
it cuts x2 + y2 = 4 orthogonally
a.b a.c b.c a.b a.c b.c
74. f (x)= ab cos x – b
76.
77. Let S x2 + y2 + 2gx + 2fy + c = 0
Given that
x
h
1 1 1 1 2 2 co s 48 48
(b – a ) 46
3 . Hence f has maximum at – and minimum at 2 2 80. Equation More over, max f = ab sin (/2 – ) +b
1 a 2 cos (
)+ 2
c = a2b + b (1– a2) + c = b + c and min f = – b + c.
_
of
_
the
_
_
_
plane
is
_
(r.n1 q1 ) ( r.n2 q2 ) 0 r.(n1 n 2 ) (q1 q2 ) _
Now , perpendicular to
and min f = – b + c.
required
_
_
_
_
_
n3 n 4
_
(n1 n 2 ).(n 3 n 4 ) 0 75.
lim (log 3 (ax 2 3x 1)) x 2
must be equal to 1
_
So 4a + 7 = 3 a = – 1
Also
le
log 3 ( x 2 3x 1) 1 lim x2 log 3 ( x 1)
RESONANCE
e
( x 1) (3 2x ) lim x 2 ( x 2 3x 1)
_
_
[n1 n 3 n 4 ] _
_
_
[n 2 n 3 n 4 ]
e 1/ 3
So independent on q3 and q4 only.
Page - 30
MATHEMATICS 81.
x 1 y 0 z 1 (1 0 2 4) 2 , 1 2 2 1 ( 2)2 22
L et C ( D ( CD
2
2
1, 2 1 2 , 3 1 3 ) a n d
1,
2
2
2, 3
2
2 2 5 Q( x, y, z ) , , . 3 3 3
3).
14
( 1
1
2
)² 4 ( 1
1 1
2
2
)² 9 ( 1
2
)² 1 4
84. (A)
Consider g(x) = sin–1 f(x) – x
1 1
L et G ( , , )
since g(0) = 0, g
4 5 21
=0 2
4 10 41
4 9 61 4 x -5 4 y -1 0 4 z -9 = = 2 4 6 5 5 9 2xyz2 = 2 = 4 3 1 1 2
there is at least one value of
L o c u s o f G is
82.
=
i.e.
........ (1) 2x2 + 2xy + 3y2 – 1 = 0 ........ (2) Homogenizing (2) with the help of (1), we get
–1 = 0
1 – ( f ( ))2
1 ( f ( ))2 for atleast one value of but may
f () =
0, 2
(B) Consider g(x) = f(x) –
2
since g(0) = 0, g
3x 6y 2x2 + 2xy + 3y2 – =0 k k2(2x2 + 2xy + 3y2) – (3x + 6y)2 = 0 ...... (3) circle described on AB as Diameter passes through (0, 0) AB will subtend right angle at (0, 0) coefficient of x2 + coefficient of y2 = 0 (2k2 – 9) + (3k2 – 36) = 0 k2 = 9 k = ± 3
i.e. f () =
2x
=0 2
2 =0
0, 2 false
(C) Consider g(x) = (f(x))2 –
since g(0) = 0, g
is
9
1 unit.
The slant height l of the cone is 2 units. Then the radius of the base of the cone is l2 1 4 1 3
Hence, the volume of the cone is
( 3 ) 2 .1 3
cubic units. Area of the circle on the plane which the rod traces is 3. Also, the centre of the circle is Q(x, y, z). Then
RESONANCE
2x
=0 2
there is at least one value of
= 2f() f () –
0, such that g () 2
2 for atleast one value of but may not be for all
The rod sweeps out the figure which is a cone. The distance of point (A(1, 0, – 1) from the plane
false
there is at least one value of
= f () –
83.
0, such that g () 2
f ( )
not be for all
3x 6y =1 k
1 1 2 4 |
or
0, such that g () 2
2 =0
f() f () =
1
true
4x 2 (D) Consider g(x) = f(x) –
since g(0) = 0, g
2
=0 2
Page - 31
MATHEMATICS there is at least one value of
0, such 2
f(x) = 1 – x6/5 f (x) = –
89. (A)
6 1 exist x (–1, 1) 5 x5
and f(1) = 0 = f(–1) Rolle’s theorem is applicable that g () = f () –
8 2
f () =
=0
x
xlim 0
(B)
8
true
2
2
xe t dt
0
1 x – ex x
x2
x 2 x e t dt = xlim 0 0 x 1 x – e
2
x(e ) e t dt 0 = xlim 0 0 0 1– e x
85.
2
third side will be parallel to bisectors of two given line Bisectors are
7x – y 3 5 2
=±
(–x – y 3)
2
x=
, are critical points 3
f(x) is increasing in
0, 3 and decreasing in
3 3 , 2
3 3 3 =–2 Hence f(0) = 0 , f = , f() = 0 , f 3 2 2 Least value = – 2, Greatest value =
b < a or
(D) For many - one, into function
RESONANCE
1
| a b |2
+
| a – b |2
=
1 sec 2 cos ec 2 2 2 4
=
1 4
2 2 1 tan 2 1 cot 2
=
1 4
2 2 2 tan 2 cot 2 tan2
A.M. G.M
2 cot 21 2 2
M(1, 2) lies on the director circle of 7x2 – 12y2 = 84
90. m3 + (2a + 5)m2 – 6m – 2a = 0 m1 + m2 + m3 = – (2a + 5) m1 m2 + m2 m3 + m3 m1= – 6 m1 m2 m3 = 2a For (A) 3
3 3 2
a+
88. (A) If function is one- one & onto then a = b since every element of set B should have exactly one pre-image in A. (B) For one - one, into function, every element of set B should have either one pre-image or no pre-image in set A. no. of elements in set B > no. of elements in A. b > a (C) For many - one, onto function, every element of set B should have one or more than one pre-images in A. n (B) < n (A)
0 1 1 =–2 –1
2 2 a – b | = 1 + 1 – 2cos = 2(1 – cos) = 4 sin /2
1
(D)
=
and |
Bisectors are 3x + y – 1 = 0 and x – 3y + 9 = 0 Now required third side will be parallel to these bisectors 3x + y + 7 = 0 or x – 3y – 31 = 0
but max {f(x), g(x)} will be non-differentiable when f(x) = g(x) no of points where f(x) = g(x) are four in [0, 2] 87. f(x) = 0 2cosx + 2cos2x = 0
2
| a + b | = 1 + 1 + 2cos= 2(1 + cos ) = 4 cos2/2
(C)
86. f(x) = cos–1 (cos 2x), g(x) = |cos x| f(x), and g(x) both are even and periodic so max {f(x), g(x)} and min{f(x), g(x)} will also be periodic and even.
2
x(e x .2x) e x e x = xlim = 0 –e x
a>b
aR & b R .
m
i =–1
i 1
a – 2a – 5 = – 1 a = – 4 m1 m2 m3 = –8 m1 = 1, m2 = –2, m3 = 4 m1 m2 m3 = – 8 for (B) a – 2a – 5 = – 5 a = 0 m1 m2 m3 = 0 m1 = 1, m2 = 0, m3 = –6 3
a+
m
i =0+0=0
(C is correct)
i 1
for (D) a + 2a = 32 not possible Hence (D) is incorrect. 91. fof(g(x)) = 1 + sin(fog(x)) f(x) = 1 + sinx
f(1 – x) = 1 + sin(1 – x)
Page - 32
MATHEMATICS 92. S1 : a, b, c must belongs to R. S3 : z =
a + ib = a – ib
S4 : a, b R
= tan(sec–1 x 0
2)=1
P(0) = 1 b = 1
93. S1 : D = 0 roots are real and equal if a, b, c R. S2 : Roots are integers S3 :
dy dx
a1 b1 c1 a 2 b 2 c 2 is true iff roots are both common
S4 : If Numerator and Denominator have common factor then Rf can be equal to Rg.
ax3 ax 2 – +1 3 2
P(x) =
3 ax 4 ax – +x+c 6 12
P(x) =
P(–1) = 1 a
1 1 12 6 – 1 + c = 1
a + c = 2 .........(1) 4
F2 P( ) F
1
S
S G
94.
and P(1) = 1 a = 12c solving (1) and (2), we get
1 and a = 6 2
Let I(h, k) and G(ae2cos, 0) c=
h=
e(a cos ) ae2 cos ae cos (1 e) = = aecos 1 e 1 e
and k =
e(b sin ) 0 be = sin 1 e 1 e
P(x) =
P(0) =
1 x4 – x3 + x + 2 2
1 2 2x3 – 3x2 + 1 = 0 (2x + 1)(x – 1)2 = 0
P(x) = 0
be sin h = aecos ; k = 1 e h = 3cos, k =
1 a = 6, b = 3 3 and e = 2
3 sin
locus of incentre I(h, k) is
and its eccentricity =
1–
99. to 100. cos–1
x2 y2 =1 9 3
3 = 9
2 3
9 3 = P 3, 2 and G(ae2cos,0) G 4 , 0 3
95 to 96 f(1.1) = f(1.2) f is may one log{x} [x] < 2 [x] > {x}2 [x] > (x – [x])2 x2 – 2[x] x + [x]2 – [x] < 0 x [–1, 0) x2 + 2x + 2 < 0, not possible x [0, 1) x2 < 0, not possible x (1, 2)
97 to 98. two points of inflection occurs at x = 1 and x = 0 P(1) = P(0) = 0 P(x) = ax(x –1) P(x) =
ax ax – 3 2
xy – 6
x2 4
1
1
x2 4
1–
1 – cos2 =
x2 4
–
1
y2 9
xy y2 cos = 6 9
xy cos = 6
2
xy cos y2 x2y2 x2y2 + = + cos2 – 3 9 36 36
x2 4
+
xy cos y2 – 3 9
9x2 + 4y2 – 12 xycos = 36 sin2 N = 36 and (cos–1 x)2 – (sin–1 x)2 > 0
cos 1 x sin1 x > 0 2
0 < 2 infinitely may solution.
3
cos =
2 2 1 x 1 y 4 9
P
15 9 8 , 4
y x + cos–1 = 3 2
1
we know that the straight lines joining SF1 and SF2 bisects the normal PG. Required point of intersection is midpoint of PG.
Required point =
.........(2)
1 1, 2
x
p=–1, q=
[p, q)
1 2
2
+b
RESONANCE
Page - 33
MATHEMATICS N–6 =
99.
36
–6=0
1 1, 2
3sin–1(sinM) = 3sin–1
x2 – (tan(3sin–1(sinM))) x + a4 = 0 x2 – (tan/4) x + 3 = 0 x2 – x + 3 = 0 + = 1, = 3 () – (+) = 3 – 1 = 2
100. sec–1x is not defined at x = 0 101. to 102.
P 4,
–26 –10 , 3 3
105. to 106.
AB = ˆi – ˆj 3kˆ , AC – ˆi – ˆj 2kˆ and BC –2iˆ – kˆ
Roots are 1,
AC BC Fourth vertex D is (4, 2, 0)
–
AB × AC = ˆi – 5ˆj – 2kˆ Equation of base is x – 5y – 2z + 6 = 0 Let E(x,y,z) be the foot of the perpendicular drawn from P to the base
x – 4 y 26 / 3 z 10 / 3 = = = r (let) –5 1 –2
(4 + r, – 5r –
It lies on the base
–
26 10 – 2r + 6 = 0 – 2– 3 3
r=–2
E 2,
4 2 , 3 3
EP = 2iˆ – 10ˆj – 4kˆ height = EP = 120
Volume =
+m+n=4
1 1 | × || |= 3 AC BC EP 3
25 1 4
120
60 = 20 cubic unit 3
=
103. to 104
2( 3 – 1) 3 2 3 f(x) = tan–1 2 x + 2 2 3 x2 + 2 + 2 3 x x x 2 2 x
2( 3 – 1) = = M which occurs at x2 = 2( 3 1) 12
fmax = tan–1
3 x
2
i.e. at x = 31/4 = a
and fmin = 0 = m at x = 0 cos–1x + cos–1y =
3 tan
–
2 a a>0
2
a + 2a + 5 (5, )
1 2 < – 4 a 0, 2 a
x2 + y2 = 4 c1(0, 0)
; ;
x2 + y2 – 24x – 10y + 2 = 0 c2 = (12, 5)
r1 = 2
;
r2 =
169 – 2 > 0
–
169 – 2
48 , 48
Since is integer = 0, ±1, ±2,....,2 ±6 Number of possible integral values of is 13 Ans. x2 + y2 = M + 2 = 3 + 2 = 5 tangent at (1, – 2)
2( 3 1)
2 1 log0.2x > 3 = log0.2(0.2)3
–1 0 0 0 1 0 So N = 0 0 3 det.(N) = – 3
T –1 (P T ) –1 P PQ P
= P–1N (PT)–1
1 125
0 < x < (0.2)3
(B)
(e x 1)(2x 3)( x 2 x 2) (sin x 2) x( x 1)
( e x 1)( x 3 / 2) 3 0 x < – 1 or x x( x 1) 2
x (–, –1)
det.(adj N) = (detN)2 = (–3)2 = 9 Ans.
112.Given PQPT = N
Tr.(P) = 3, Tr. (P3) = 27 = 33 Tr.(P5) = 243 = 35 Tr. (P2011) = 32011 Also Tr.(P2) = 11 = 32 + 2 Tr.(P4) = 34 + 2 Tr.(P2012) = 32012 + 2 Ans.
0 0 t=a+3 or t = a2 + 2 0a+35 or 0 a2 + 2 5 –3 a 2 or –2 a2 3 –3 a 2
or –
3 a 3
acceptable integer values of a’ are –3, –2, –1, 0, 1, 2 6 is the answer 127. Centre of circles c1, c2, c3 are in A.P. General term for abscissa of centres = 1 + (n – 1).3 = 3n – 2 centre of c5 is (13, 0) Radius of circles are in G.P. Rn = 1.2n – 1 = 2n – 1 R3 = 4 and centre of c3 is (7, 0) tangents of circle c3 intersect each other at (13, 0) equation of any line passing through (13, 0) is y – 0 = m(x – 13) mx – y – 13m = 0 now it will be required tangents if
7m – 0 – 13m m2 1
=4
2 36m2 = 16m2 + 16 20m2 = 16 m = ±
5
Page - 37
MATHEMATICS
2 Let m1 =
f (2) = – 4e–4 – 4e–4
2
5
, m2 = –
f (2) is negative hence at a = 2
5
f (a) is maximum
10|m1m2| = 8
M1 M2 M3 L1 L2 128.
M4
L3 L4
M5
M6
4
Required probability =
C1 3 C 2 6
C2
4 = 5
1 4 =1–1× = 5 5
1 and minimum value of 2
minimum value of 2x2 – 2x + 1 is
Alter : Required probability = 1 – (no languages is common)
sin2x is 0 minimum value of
e( 2 x
2
– 2 x 1) sin2 x
is 1
135. a = b 136. sin 3x = cos 4x
129. Q1(x) 0 a > 0 and b2 – 4ac 0 Q2(x) 0 b > 0 and c2 – 4ab 0 Q3(x) 0 c > 0 and a2 – 4bc 0
cos4x = cos
3 x cos4x = sinx ± 3 x 2 2
137. T r = cot –1 (2r2)
2
a – 4( ab) 0
2
a 4( ab)
2
1 1 134. 2x – 2x + 1 = 2 x + 2 2 2
36 – 4k2 > 0
133. f'(x) has D > 0 k (–3, 3) Ans. 3
a > 0, b > 0 and c > 0
S=
=
a ........(1) ab 4
=
( 2r 1) – ( 2r – 1) 1 ( 2r 1)( 2r – 1)
–1
tan
–1
( 2r 1) – tan –1( 2r – 1)
r 1
= (tan–13 – tan–11) + (tan–15 – tan–13) + (tan–17 – tan–15) +.... + tan–1
2
a > 1 .........(2) ab
from (1) and (2), we can say that
S=–
a2 ab 4
+ = 4 2 4
5 x.2 = x 2 1
= 2(–x2 + x) e–2x f (x) = 0 at x = 1 and f (x) > 0 in (0, 1) and f (x) < 0 in (1, ) f(x) is maximum at x = 1
x = 1 only
1 f(1) =
131. f(x) = 2x³ – 3 (k + 2)x² + 12kx – 7, – 4 k 6, k l f'(x) = 6(x² – (k + 2)x + 2k) = 6(x – k) (x – 2)
2
e
=
2
3 2–
Equation of AC
e – ( x – 2 ) dx
a–2
y–0= 2
(2 – ) (x – ) 3
2
e –(a) – e –(a – 4) = 0
2
1 = e2
[–1] = 7 139. Slope of AB =
a 2
For f(x) to be invertible, k must be 2 only.
f (a) =
1 = k k = 4 4
138. f (x) = –2x2 . e–2x + 2x . e–2x
130. log10 x
Given S =
[k] = 0
possible of integers in the range are 2,3,4 Number of integers in the range is 3.
132. Let f(a) =
1 2 2r
–1
we know that (a – b)2 + (b – c)2 + (c – a)2 > 0
1<
tan r 1
a2 – ab 0
tan r 1
2
r 1
ab 0
cot –1( 2r 2 )
2
e – a – e – ( a – 4 ) – a2 = – a2 – 8a + 16
a = 2 2
f (a) = = – 2a e
e – a (–2a)
– a2
+ 2(a – 4)
2
–
e – ( a – 4 ) (–2(a – 4))
e –(a – 4)
2
C 0,
RESONANCE
( 2 – ) 3
Page - 38
MATHEMATICS Let M(h,k) a = 2h, 2k =
( 2 – ) 3
6k = 2h(2 – 2h) 3y = x(2 – 2x) 2x2 – 2x + 3y = 0 a = 2, b = 3 a+b=5
1
144. Let =
0
=
=
20 2 20 5 20 8
[] = 2 Ans.
142. Using the idea of the differentiation of determinant, we get
cos( x ) cos( x ) f(x) =
cos( x )
cos( x ) cos( x )
cos( x ) + cos( – ) cos( – ) cos( – )
sin( x )
sin( x )
sin( x )
– sin( x ) – sin( x ) – sin( x ) cos( – ) sin( x )
cos( – ) sin( x )
cos( – ) sin( x )
cos( x ) cos( x ) cos( x )
+
0
0
=0+0+0 f(x) = 0 for all x f(x) = a constant But f(9) =
0
9
f(x) = for all x.
0
dx
sin–1 1– x x2 – x 1
dx
......(2)
– x 1
1– x 2
sin–1 x = cos–1
(1) + (2) we get sin–1
1
sin–1 x cos –1 x
x2 – x 1
1
4 =
1
0
1– x = cos–1 x
2
1 3 x – 2 2
2
dx
dx
1
1 –1 2x – 1 tan 3 = 4 3 0 2
=
n = 108
n2 6 6 = 2 3 108 2 3 3
n = 4 Ans. 27 145. K r r a b ..........(i) pre cross (i) with a both side K(a r ) a ( r a ) a b 2 K ( a r ) | a | r – ( a . r ) a a b 2 K(K r – b) | a | r – (a . r ) a a b (using equation (i))
r (K 2 | a |2 ) a b Kb (a . r ) a ..........(ii) Dot (i) with a both side K(a . r ) a . b a.b a. r ..........(iii) K
by (ii) & (iii)
f(k) k 1
2
0
20 3 1 1/ 2 1/ 3 20 3 (20) 1 4/5 2/3 6 = 47 1 7/8 7/9 20 9
50 21
=
x
1
dx
– (1– x) 1
cos –1 x
0
0 –3 /10 –1/ 3 (20)3 0 –3 / 40 –1/ 9 = 47 1 7/8 7/9
2
2=
applying R1 R1 – R2 and R2 R2 – R3
(1– x) 1
1 20 141. an = and d = 20 n
20 = 4 20 7
sin –1 1– x
0
......(1)
x 1
140. x2 + (a – b) x + (1 – a – b) = 0 D>0 (a – b)2 – 4 × 1 × (1 – a – b) > 0 a2 + b2 – 2ab – 4 + 4a + 4b > 0 b2 + 2b (2 – a) + (a2 + 4a – 4) > 0 4(2 – a)2 – 4 × 1 × (a2 + 4a – 4) < 0 4 + a2 – 4a – a2 – 4a + 4 < 0 8a – 8 > 0 a>1
20
sin x dx 2 – x 1 –1
=9
r
143. Put log10x = t [t2 + t] = 2t + 1 2t + 1 t2 + t < 2t + 2 and 2t t
RESONANCE
a .b Kb a a b K 2 | a |2 K 1
mn=3
Page - 39
MATHEMATICS 146. For defining f(x)
(16 – x), (20 – 3x) N ; (2x – 1), (4x – 5) W
f(x) =
(16 – x) (2x – 1) ; (20 – 3x) (4x – 5) Hence x = 2, 3 x = 2, 3 when x = 2, f(x) = 14C3 + 14C3 = 728 when x = 3, f(x) = 13C5 + 11C4 = 1617 Hence sum 728 + 1617 = 2345
F(x) xF'(x) dx
f(x) = x F(x) + c
f
2 = f = F + c 2 2 2 2 4
147. z = (z – z2 + 2z3)(2 – z + z2) = ( – 2 + 2) (2 – + 2 ) = |2 – + 2 |2 =
f(x) = xF(x) f() = x F() = ()0 |cos(f())| = |cos | = 1
148. On expanding the determinant, we get 7sin3 + 14cos2 – 14 = 0 sin3 – 2(1 – cos2) = 0 3sin – 4sin3 – 4sin2 = 0 sin (2sin – 1)(2sin + 3) = 0 sin = 0 or sin =
and for cos
hence f (1) < 0 3 – 3 sin – 2 cos2 < 0
1 < sin < 1 2
In [0, 2],
152. [x + a] = [x] + a ; if a I
5 solutions
1 lies between the roots
1 2
Minimum if 'a' for sin
5 = 0, , 2, , 6 6 149.
1
2 = 2 +cc=0 2 4
| 2 – i 3 |2 = 7
2n +
2 sin2 – 3 sin + 1 < 0
5 0, x
6, 3
g , g 6 3
g , g 6 3
(1 + xcosx . nx + sin x)dx
sin x sin x x.x sin x cos x ln x x x dx
i.e.
1 3 , 2 3 3 2
Let F(x) = xsinx range of f(x) is F’(x) =
2
4 4 x < 3 9
g(x) is strictly increasing
3 cos3 2
x
–
(R1 R1 + R2 + R3)
151. f(x) =
4 4 2 0 for all x , 6 3
g(x) is monotonically increasing
3 = 2 sin + tan – . +1 6 6 6 6
f(9801) = 99
1 168. First note that (3, 3) is in the interior of the arc, as
RESONANCE
=1+
3
–
1 +1 2
Page - 43
MATHEMATICS
f
3 1 = n 3 6 2
f
= n 3
a + b = n
3
=
3 1 + n 2 3 2 3
1 0 For g(x) to be strictly increasing f (2x2 – 1) – f (1 – x2) > 0 f(2x2 – 1) > f(1 – x2) 2x2 – 1 > 1 – x2 3x2 > 2
x>
2 3
( x > 0) Case–II x < 0 For g(x) to be strictly increasing f (2x2 – 1) – f (1 – x2) < 0 f (2x2 – 1) < f (1 – x2) 2x2 – 1 < 1 – x2 3x2 < 2 x2 <
2 2 – f f 2000 1999 f(x + 1) > f(x – 1)
or,
2 cos x + 1 = 0
173. and increases on (a1 + a2 + a3 + a4) = 21, a5 + a3 + a6 + a7 = 21, a1 + a2 + a3 + a4 + a5 + a6 + a7 = 35 a3 = 7, a1 + a2 + a4 = 14, a5 + a6 + a7 = 14
RESONANCE
2 3
or, cos x + 2 = 0 Sign scheme for f(x) in [0, ] is as below.
f(x) decreases on
, 2
(not possible)
2 0, , 2 3 2 , 2 3
Page - 44
MATHEMATICS 177. Number of selections of 7 digits out of the digit 1,2,3,.....,9 = 9c7 Number of digits out of these 7 selected digits excluding the greatest digit = 6 These 6 digits can be divided in two group each having 3 6!
digits 3 ! 3 ! 2!
6
C3 x
179. (n3 – 1)3 = (n + 1)3(n – 1)3 (n + 1)3 – (n – 1)3 = 6n2 + 2
3n2 1 2
(n – 1)
1
178.
6
C7 . C 3 .
2 2!
9
3 3 2 1 6n 2 1 (n 1) – (n – 1) = 3 3 2 (n2 – 1)3 2 (n 1) (n – 1)
2!
But the 3 digits on one side can go on the other side Required number of ways = 9
3 =
6
C 7 . C3
9
=
6
1 1 (n – 1)3 – (n 1)3
S=
C2 . C3 = 720
b 40 12
1 2
1 2
1 1 1 1 1 1 1 1 3 – 3 3 – 3 3 – 3 3 – 3 ... 6 1 3 2 4 3 5 4
c 3 10
cos1 cos x x a b c
=
9 1 1 1 9 = 16S = 9 16 = S 2 8 16
cos1 cos x x 1/n
(2n 1)(2n 2)(2n 3).......(2n n) A 180. = lim n B (n 1)(n 2)(n 3).........(n n)
ln (A/B)
r 2 n ln r 1 1 r / n n
1 lim = n n
Integral value of x satisfying given equations = 2,3,4 1
0
(2 x) 27 ln dx = 16 (1 x)
A = 27
RESONANCE
&
B = 16
Page - 45
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Main Campus Pragati Building, 2nd Floor, Sanskrutik Sankul Building, Jhansi Rani Square, Sita Buldi, Nagpur - 440012 Tel: 0712 - 3017222, 3192222, 6060660. Mob.: 9372137857, 9373507997 / 8 e-mail:
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University Road Campus A-One School, University Road, Udaipur-1 Tel.: 0294-3192222, 2427858,
Extension Campus Vinamra Building, Above PNB Bank, Beralia House. Plot No. 91, Chaprunagar Square, Central Avenue Road, Nagpur - 440 008 Tel: 0712-6462622, 3922333, 6060661
Plot No-21/5, Near Regional Science Centre, Acharya Vihar, Bhubneshwar-751013 Tel.: 0674-3192222, 3274919, 6060660/1 e-mail:
[email protected]
BHUBANESWAR
MUMBAI
AHMEDABAD
Andheri Campus (Regional Office) Spenta Mansion, 2nd Floor, Above Moti Mahal Hotel, Opp Station, Andheri (W), Mumbai - 400 058 Tel.: 022-3192 2222, 60606600 e-mail:
[email protected]
2nd Floor, C-Block, Galaxy Bazaar, Opp. Sunrise Park, Near Himalaya Mall, Off Drive-In Road Vastrapur, Ahmedabad Tel.: 079-31922222/3, 60606600/1/2 e-mail:
[email protected] BRANCHES: SHAHIBAUG I MANINAGAR I GANDHINAGAR
Borivali Campus 2nd Floor, Raghuvir Tower, Junction of SVP Road & Ganjawala Lane, Chamunda Circle, Borivali (W), Mumbai Tel.: 022-31922226, 6060 6601 e-mail:
[email protected] Churchgate Campus Jai Hind College, Admissions Office "A" Road, Churchgate Mumbai-400020 (Maharashtra) Tel.: 022-31922223, 60606602 e-mail:
[email protected] Dadar Campus C 8 (A) Kasturchand Mill Compound, Ground Floor, Near Manish Market, Senapati Bapat Marg, Near Dadar west Railway station,Dadar (W), Mumbai Tel.: 022-31922224 / 9769210458 e-mail:
[email protected] Govandi Campus 1st Floor, Runwal Complex, Opp ICICI Bank, Govandi Station Road, Govandi (E), Mumbai Tel.: 022-31922225, 9930269183 e-mail:
[email protected] Nerul Campus C-11, Amrita Sadan, Plot no. 13/14, Sector - 22, Oppo. Nerul Rly. Station, Nerul (W), Navi Mumbai-6 Tel.: 022-31922227, 60606605 e-mail:
[email protected] Panvel Campus Deep Dhara, Plot No.08, Road No.18, Sector-1, Opp Geetanjali Building, New Panvel - 410 206 Tel.: 022-31922232, 60606606 e-mail:
[email protected]
LUCKNOW
Thane Campus Paranjpe Udyog Bhavan, 2nd floor, Above Khandelwal Sweets, Opp. station Thane (W) - 400607 Tel.: 022-31922228, 32191498, 6060 6607 e-mail:
[email protected]
Hazaratganj Campus 13/1, Moti Chamber, Rana Pratap Marg, YMCA Complex, Ground Floor, In Front of Gauri Apartment, R.F. Bahaduri Marg, Hazratganj, Lucknow (U.P.) - 1 Tel.: 0522-3192222, 3205854, 6060660/1 Mob.: 9369225676 e-mail:
[email protected]
Kharghar Campus 5/6/7/8, Swami Samarth Bldg., Sector-12, Opp. Kendriya Vihar Colony, Behind Syndicate Bank, Behind Mercantile Bank, Kharghar Tel.: 022-32191499, 60606608 e-mail:
[email protected]
PATNA Kankarbagh Campus NC/8, Gayatri Mandir Road, Kankar Bagh, Patna (Bihar) - 800020 Tel.: 0612-3192222, 2350070 e-mail:
[email protected] Boring Road Campus 3rd Floor A G Palace, Besides St. John's High School, Near Panchmukhi Hanuman Mandir, East Boring Canal Road, Patna Tel.: 0612-3192223, 2520109 e-mail:
[email protected]
JODHPUR Aanchal Complex (Basement), Near Manidhari Hospital, Residency Road, Sardarpura, Jodhpur Tel.: 0291-3192222, 6060660 e-mail:
[email protected]
AJMER Plot No. 1408, Aryan College Building, Near Ram Bhawan, Shastri Nagar, Ajmer-305001 Tel.: 0145-3192222, 2621614, 6060660 e-mail:
[email protected]
INDORE 1st Floor, Gravity Tower, Opp. Agrawal Stores, Indore - 452003 (MP) Tel: 0731-3192222, 4274200 e-mail:
[email protected]
SIKAR Vishwakarma Apartment, Piprali Road, Sikar-332001 Tel: 01572-319222, 606066 & 258222 e-mail:
[email protected]
AGRA 1, Kailash Vihar, Behind St. Conrad's college, Khandari, By-Pass Road, Agra-282005 Tel: 0562-3192222 e-mail:
[email protected]
RANCHI
ALLAHABAD
GWALIOR 1st & 2nd Flr, Bansi Plaza, Near S.P. Office, City Centre, Gwalior - 474012 Tel.: 0751-6060660, 9669688934 e-mail:
[email protected]
JABALPUR 1st Floor, Nayan Tower, Near Madan Mahal Police Station, Shivaji Chowk, Napier Town - 2 Tel.: 0761-6060660, 8120005717 e-mail:
[email protected]
AURANGABAD 2nd Floor, Alaknanda Building, Near Baba Petrol Pump, Adalat Road Aurangabad - 431001 Tel.: 9910066021 e-mail:
[email protected]
RAIPUR 1st & 2nd Floor, Vandana Auto Building, Opp Raj Kumar College, Raipur Pin code- 492001 Tel.: 0771-6060660, 7771007840 e-mail:
[email protected]
NASHIK Mangal Plaza, Second Floor, Near Kalika Mandir, Old Agra Road, Nashik-422002 Mob.: 09730877804 e-mail:
[email protected]
CHANDRAPUR Resonance Study Centre, Zanzari Complex, Z.P. Road, Near Hotel Rasraj, Chandrapur-442401
SURAT 21st Century Business Centre, Near World Trade Centre, Udhna Darwaja, Ring Road, Surat Tel.: 7874438681, 09725552181 e-mail:
[email protected]
RAJKOT 2nd Floor, Gayatri Commercial Complex, Opp. Parimal School, Kalavaad Road, Rajkot-360005 Tel.: 09314805678, 09974802362 e-mail:
[email protected]
VADODARA Royal Hub, Next to Seven Seas Mall, Fatehgunj, Vadodara - 390002 Tel.: 0265-6060660, 7567810117 e-mail:
[email protected]
VISAKHAPATNAM (ICCP) Vikas Vidyaniketan, Door No.: #1-1-1/1, Sheela Nagar, NH-5, Visakhapatnam, (AP) - 530012 Tel: 0891-2757575 e-mail:
[email protected]
BHAVNAGAR (ICCP) Sardar Patel Educational Institute, Kalvibid, Bhavnagar 364002 (Gujarat) Mob.: 8000816767 / 8000716868
DEHRADUN (ICCP) SelaQui International School, Chakrata Road, Dehradun - UK, INDIA Tel: 91-135-3051000 | Mob.: 9873743433
NANDED Beside Akola Urban Cooperative Bank, V.I.P Road, Paras Square, Nanded.- 431602 Tel: 02462-250220, 606066 e-mail:
[email protected]
CORPORATE OFFICE: CG Tower, A-46 & 52, IPIA , Near City Mall, Jhalawar Road, Kota (Rajasthan) - 324005 Reg. Office: J-2, Jawahar Nagar Main Road, Kota (Raj.) - 324005 | Tel. No.: 0744-3192222, 3012222, 3022222 | Fax : 022-39167222, 0744-2427144
To know more: sms RESO at 56677 | Toll Free : 1800 200 2244 | Website: www.resonance.ac.in | e-mail:
[email protected]
